How to ensure when the derivative approaches zero, the integral approaches a constant?

Question

I asked a very similar question here. But now this is different. Suppose $f(t)$ is differentiable and $c$ is a finite constant, then the following statement looks correct, but in fact it is not: \begin{equation} \lim\limits_{t \to \infty} f'(t) = 0 \implies \lim\limits_{t \to \infty} f(t)=c \end{equation} A counter-example is $f(t)=\ln(t)$. Now the question is, what condition should be used for the above statement to be true?

The inverse implication is always true: $$ \begin{equation} \lim\limits_{t \to \infty} f(t)=c\implies \lim\limits_{t \to \infty} f'(t) = 0 \end{equation}$$ Thus, we find that $\lim\limits_{t \to \infty} f'(t) = 0 $ is a necessary condition for the function being convergent at $x\to\infty$. I think that what you're actually looking for is a statement $A$ so that $$\left( \lim\limits_{t \to \infty} f'(t) = 0\right) \land A \implies \lim\limits_{t \to \infty} f'(t) = 0 $$ is true, but so that the formula $$ A \implies \lim\limits_{t \to \infty} f'(t) = 0 $$ doesn't hold in general. — Sudix, Jul 24 '19 at 15:32
The inverse implication is NOT true. See the link in my question. — winston, Jul 27 '19 at 21:01

score 3 · Answer 1 · answered Dec 18 '18 at 16:24

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An equivalent condition is that $$ \int_a^\infty f'(t)\,dt $$ converges as an improper Riemann integral for some $a\in\Bbb R$. This happens for instance if $|f'(x)|\le C\,x^{-p}$ for some $C\ge0$ and $p>1$.

answered Dec 18 '18 at 16:24

Julián Aguirre

76,354

Is there any proof of your result? – winston Dec 18 '18 at 16:39
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$$f(x)=f(a)+\int_a^xf'(t),dt,\quad x>a.$$ – Julián Aguirre Dec 18 '18 at 16:40
True if $f'$ is Riemann integrable on bounded intervals. – zhw. Dec 18 '18 at 17:14
Does it mean that $\lim\limits_{t \to \infty} f'(t)=0$ does not make any difference to the condition? Doesn't it somehow relax your condition in some way? – winston Dec 18 '18 at 20:21
If $\lim_{x\to\infty}f'(x)$ exists, it must be $0$. But it may happen that the limit does not exist. – Julián Aguirre Dec 19 '18 at 08:28

score 0 · Answer 2 · answered Jul 26 '19 at 03:23

One way you could potentially change your statement so that it is true is to assert that the function is bounded asymptotically (as $t \rightarrow \infty$).

If $\lim\limits_{t \rightarrow \infty} f'(t) = 0$ and $\exists n,M$ such that $\forall t > n, |f(t)| \le M$, then $\exists c$ such that $\lim\limits_{t \rightarrow \infty} f(t) = c$

I don't have a proof of this currently, but I believe it should be true, and in fact I believe that this also works in reverse. I will provide a proof if I can come up with one.

score 0 · Answer 3 · answered Jul 26 '19 at 04:43

Two ways for a differentiable (and hence, continuous) function $f(t)$ to satisfy $\lim_{t \to \infty} f(t) = c$:

(1) $f(t) = c$; i.e., is a constant function

(2) $f(t)$ has a horizontal asymptote $y=c$ in the positive x-direction.

Hence, restrict $f(t)$ to be either of these two function types and your original assertion $\lim_{t \to \infty} f'(t) \Rightarrow \lim_{t \to \infty} f(t)$ will be true.

How to ensure when the derivative approaches zero, the integral approaches a constant?

3 Answers3